Integrals Involving Trigonometric Substitution
I. Integrals of the Form
a2  x2
a 2  x 2 where a is a positive
constant. If you make the substitution x  a sin  , where       , then
2
2
Suppose that you have an integral of the form
a 2  x 2  a 2  a 2 sin 2   a 2 (1  sin 2  )  a 2 cos 2   a cos  because
cos   0 since   2     2 . Also, dx  a cos  d . Also, x  a sin  
x
 x
sin      arcsin   .
a
a
x
a
s
sin    y  s and r = a 

a
a2  x2
a2  s2
.
a
x 2  s 2  a 2  x  a 2  s 2  cos  
Example: Find

4  x 2 dx .
Let x  2 sin  where       . Then
2
2
2 cos d .

4  x 2  2 cos and dx 


4  x 2 dx  (2 cos )(2 cos d )  4 cos 2 d 
1
1

4    sin  cos   C  2  2 sin  cos  C =
2
2

x
2
 x   x  4  x
2 arcsin    2 
2
 2   2 

C =



2
 x x 4 x
2 arcsin   
+ C.
2
2
Let’s check this answer.
2
4  x2
2
d 

 x x 4 x
2
arcsin




dx 
2
2

1







1
2
 1 x
2

 
2

1  1    2 x
    x
2  2   2 4  x 2



2
 x2
  4  x 2  




4  x2 2 4  x2

4  x 2 4  x 2  4  x 2 2(4  x 2 )


 4  x 2 and so it checks.
2
2
2
2 4 x
2 4 x
a2  x2
II. Integrals of the Form
a 2  x 2 where a is a positive
constant. If you make the substitution x  a tan  , where       , then
2
2
Suppose that you have an integral of the form
a 2  x 2  a 2  a 2 tan 2   a 2 (1  tan 2  )  a 2 sec 2   a sec  because
sec  0 since   2     2 . Also, dx  a sec 2  d . Also, x  a tan  
x
x
tan      arctan   .
a
a
a2  x2
x
tan  
s
 y  s and x = a 
a

a
a s r r  a s .
2
2
2
Example: Find
2

2
x 2  4 dx .
Let x  2 tan  where       . Then
2
2
2 sec 2  d .

x 2  4  2 sec and dx =


x 2  4 dx  (2 sec )(2 sec 2 d )  4 sec 3 d 
1
1

4 sec  tan   ln sec   tan    C  2 sec  tan   2 ln sec   tan   C =
2
2

 x 2  4  x 
   2 ln
2

 2 
2


x x2  4
 2 ln
2
x2  4 x
 C 
2
2
x2  4
x

x2  4  x  C .
2
2

x2  4  x  


 2x

1  2x  1 2
1
x2
x

x  4  2
 1 




2
2
2  x2  4  2
x2  4
 x  4  x  2 x  4 

 x  x2  4  x2  x2  4  4
x2  4
1
2x 2  8
 2






2
2
x 2  4 
2 x2  4
2 x2  4
 x  4  x  
x2  4
 x 2  4 and so it checks.
2
x 4
Let’s check this answer.
d 1
2
 x x  4  2 ln
dx  2

III. Integrals of the Form

x2  a2
x 2  a 2 where a is a positive
constant. If you make the substitution x  a sec , where 0     or     3 ,
2
2
Suppose that you have an integral of the form
x 2  a 2  a 2 sec 2   a 2  a 2 (sec 2   1)  a 2 tan 2   a tan  because
then
tan   0 for 0     2 or     3 2 . Also, dx  a sec tan  d . Also,
x
 x
x  a sec   sec      arc sec  .
a
a
s
Consider sec   where 0      s  0 
2
a
x = a and r = s  a 2  y 2  s 2  y  s 2  a 2 .
s
Consider sec   where     3  s  0 
2
a
x2  a2

x  a and r  s  a 2  y 2  s 2  y   s 2  a 2 .
Example: Find

a
x2  4
dx .
x
Let x  2 sec where 0     or     3 . Then
2
2
dx  2 sec  tan  d . Thus,

x


x2  4
dx 
x

x 2  4  2 tan  and
2 tan  2 sec tan  d  
2 sec 
2 tan 2  d  2 (sec 2   1) d  2 tan   2  C 
 x2  4 
  2arc sec x   C =
2

2 
2

x2  4
x

3
2
 x
x 2  4  2arc sec   C . This answer does check.
2
a 2  u 2 , a 2  u 2 , and
IV. Integrals of the Form
u2  a2
A. Sometimes, you should make a u–substitution before you make a trigonometric
substitution.
Example: Find

9  4x 2
dx .
x2
Let u 2  4 x 2  u  2 x  du  2dx . Also, a 2  9  a  3 .

9  4x 2
dx = 2
x2
where  
2

 
9  4x 2
 2 dx  2
4x 2
2
.

9  u2
du . Let u  3sin 
u2
9  u 2  3 cos and du  3 cos d 
3 cos (3 cos d )
 2 cot 2  d 
(3 sin  ) 2



2 (csc   1) d  2 cot   2  C 

2
9  u2
du  2
u2
2
 9  u2 
  2 arcsin  u   C 
 2
 u

3


 9  4x 2 
  2 arcsin  2 x   C =
 2
 2x 
 3 


 9  4x 2
 2x 
 2 arcsin    C .
x
 3 
u
3

9  u2
B. Sometimes, you have to complete the square before making a u–substitution or a
trigonometric substitution.
Example 1: Find

1
dx .
4 x  8 x  13
2
Complete the square for 4 x 2  8 x  13 :
4 x 2  8 x  13  4( x 2  2 x )  13
= 4( x 2  2 x  1)  13  4
= 4( x  1) 2  9
4
Let u 2  4( x  1) 2  u  2( x  1)  du  2dx . Also, a 2  9  a  3 .
1
1
1
1
dx 
dx 
du 
Thus,
2
2
2
2 u 9
4 x  8 x  13
4( x  1)  9



1 1
1
 u 
 2( x  1) 
C .
 arctan    C  arctan 
2 3
6
 3 
 3 
Example 2: Find

15  6 x  9 x 2 dx .
Complete the square for 15  6 x  9 x 2 :
15  6 x  9 x 2  9 x 2  6 x  15
2


 9 x 2  x   15
3


2
1

 9 x 2  x    15  1
3
9

2
 16  (9 x  6 x  1)  16  (3x  1) 2
Let u 2  (3x  1) 2  u  3x  1  du  3dx . Also, a 2  16  a  4 .
1
Thus,
15  6 x  9 x 2 dx 
16  (3x  1) 2 3dx 
3
1
16  u 2 du . Let u  4 sin  where       . Then,
2
2
3
1
16  u 2 du =
16  u 2  4 cos and du  4 cos  d .
3
1
16
8
8
(4 cos )(4 cos d ) 
cos 2 d    sin  cos  C 
3
3
3
3





2
8
 u  8  u  16  u
arcsin     
3
4
 4  3  4 


C 


2
8
 3x  1  (3x  1) 15  6 x  9 x
arcsin 

C .

3
6
 4 
u
4

16  u 2
5
Practice Sheet for Trigonometric Substitution
(1)

(2)

(3)

(4)
x2
25  4 x 2
dx =
x2
dx =
x 2  2 x  10
1
x

2
dx =
2
4x  9
1
2
x  2x  5
dx =
3
(5)

1
15  2 x  x
1
dx =
2
3
(6)

9  x 2 dx =
3
(7)
x
(8)

(9)

(10)

1
3
x2  4
dx 
x 2  1 dx 
e x 1  e 2 x dx 
x
7  6x  x
2
dx 
6
(11)

u2  a2
du 
u2
(12)

9 x 2  25
(13)

x2 1
dx 
x
(14)

(15)

(16)
 1  x 
x2
x3
1 x2
1
x 1
2
dx 
dx 
dx 
1
2 2
dx 
Solution Key for Trigonometric Substitution
(1) First, let u 2  4 x 2  u  2 x  du  2 dx 
1
8

4x 2
25  4 x 2
2dx 
1
8

u2
25  u 2

x2
25  4 x
2
dx =
du . Next, make the trigonometric substitution
5
u  5 sin   du  5 cos  d and
25  u  5 cos 
2
u

25  u 2
7
1
Thus,
8
result


u2
25  u 2
du 
1
8

25 sin 2 
5 cos  d   25 sin 2  d . Using the
5 cos 
8

1
1
sin 2 d    sin  cos  , we get
2
2

x2
25  4 x 2
2
25
25
 u  25  u  25  u
sin  cos   C 
arcsin     
16
16
5
 5  16  5 
25  2 x  25  4 x 2
 
16  5 
5
(2) By long division,
Thus,

dx 
25

16

  C  25 arcsin  2 x  

16
 5 


  C  25 arcsin  2 x   1 x 25  4 x 2  C .

16
 5  8

x2
2 x  10
2x  2
8
.
 1 2
 1 2
 2
2
x  2 x  10
x  2 x  10
x  2 x  10 x  2 x  10

x2
dx  1 dx 
x 2  2 x  10


2x  2
1
dx  8
dx 
x  2 x  10
( x  1) 2  9
2
8
 x 1
x  ln x 2  2 x  10  arctan 
C.
3
 2 
(3) First, let u 2  4 x 2  u  2 x  du  2 dx 
2
 4x
1
2
4x  9
2
2 dx  2
u
1
2
u2  9

1
x
2
4x  9
dx =
du . Next, make the trigonometric
substitution u  3 sec   du  3 sec x tan x dx and
u2  9
2
u

3
8
u 2  9  3 tan  
Thus, 2
u
1
u2  9
2
du  2

1
3 sec  tan  d   2
9
9 sec  (3 tan  )
2
2
2  u 2  9 
2  4 x 2  9 
sin   C  
C  
C 



9
9  u
9
2
x



 cos d 
4x 2  9
C .
9x
(4) First complete the square to get x 2  2 x  5  ( x  1) 2  4 . Next, let u  x  1 
du  dx 

1
x 2  2x  5
dx =

1
u2  4
du . Make the trigonometric
substitution u  2 tan   du  2 sec 2  d and
u 2  4  2 sec  
u2  4
u

2

ln
1
u 4
2
du =




1
2 sec 2  d  sec  d  ln sec   tan   C =
2 sec 
u2  4 u
  C  ln
2
2
u 2  4  u  C  ln
x 2  2 x  5  ( x  1)  C .
(5) First, complete the square: 15  2 x  x 2   x 2  2 x  15  ( x 2  2 x  1)  16 
16  ( x  1) 2 . Next, let u  x  1  du  dx 

1
16  u 2

1
15  2 x  x 2
dx 
du . Make the trigonometric substitution u  4 sin   du  4 cos  d
9
and 16  u 2  4 cos 
4
u

16  u 2
Thus,

1
16  u
2
 4 cos 4 cos d   1d    C  arcsin  4   C 
 x 1
arcsin 
C 
 4 
3

 x 1
1
dx  arcsin 
  arcsin   
2
2
 4 1
15  2 x  x

6
3
1
1
arcsin 0 
u
1
du =
. [Note: You would not need to make the trigonometric substitution
if you recognized that

u
du = arcsin    C .]
4
16  u 2
1
(6) Make the trigonometric substitution x  3 sin   dx  3 cos  d and
3 cos 
3
x

9  x2
Thus,
9
2


9  x 2 dx 


(3 cos  )(3 cos  d )  9 cos 2 d 
1  cos 2  d  9   9 sin 2  C  9   9 sin  cos  C 
2
4
2
9
 x  9  x  9  x
arcsin     
2
3
 3  2  3 
2
2
2

  C  9 arcsin  x   x 9  x  C 

2
2
3

10
9  x2 
3

9 x
3
2

9
 x x 9 x
dx =  arcsin   
2
3

2
2
3
3
9    9    9
. [Note:
     
2 2  2 2  2
of radius 3,

9
9

  arcsin 1  arcsin( 1) 
2

 3 2

9  x 2 dx represents the area of a semicircle
3
1
9
.]
 (3) 2 
2
2
(7) Make the trigonometric substitution x  2 sec   dx  2 sec  tan  d and
x 2  4  2 tan  
x
x 4
2

2
Thus,
x
1
3
x 4
2
dx 
 8 sec  2 tan  
1
3
 2 sec  tan  d   1
 cos  d 
2
x2  4
C .
8x 2
1
1
1
 x
  sin  cos   C  arc sec  
16
16
16
2
(8)
8
Make the trigonometric substitution x  tan   dx  sec 2  d and
x 2 1 
sec 
x2 1
x

1



1
(sec  )(sec 2  d )  sec 3  d  sec  tan  
2
1
1
1
ln sec x  tan x  C  x x 2  1  ln x  x 2  1  C .
2
2
2
Thus,
x 2  1 dx 
11
(9) Make the u –substitution u  e x  du  e x dx 
e
x
1  e 2 x dx 
 1  u du .
2
Next, make the trigonometric substitution u  sin   du  cos  d and
1
1
1  u 2  cos  
1  u 2 du  cos 2 d    sin  cos   C 
2
2


1
1
arcsin u  u 1  u 2  C 
2
2
1
1
arcsin e x   e x 1  e 2 x  C .
2
2
u
1

1 u2
(10) First, complete the square: 7  6 x  x 2   x 2  6 x  7  ( x 2  6 x)  7 
 ( x 2  6 x  9)  7  9  16  ( x  3) 2 . Next, let u  x  3  du  dx and
x
u 3
u
dx 
du 
du 
x  u 3
2
2
7  6x  x
16  u
16  u 2
1
3
1
1
du  
16  u 2 2 (2u du)  3
du 
2
2
2
16  u
16  u









1
1
u
u
216  u 2  2  3 arcsin    C   16  u 2  3 arcsin    C 
2
4
4
 x  3
 7  6 x  x 2  3 arcsin 
C .
 4 

(11) Make the trigonometric substitution u  a tan   du  a sec 2  d and
u 2  a 2  a sec  


(a sec  )( a sec 2  d )

a 2 tan 2 

sec 
d 
sec 2   1

3
sec  d 


u2  a2
du 
u2
u2  a2
u

sec  

 sec  
 d 
sec 2   1 

a
sec 
d  sec  d  csc  cot  d  ln sec   tan  
tan 2 


12
csc   C  ln
u2  a2 u
u2  a2
 
 C  ln
a
a
u
u2  a2
u a u 
C.
u
2
2
(12) Let u 2  9 x 2  u  3x  du  3dx  using Problem


9 x 2  25
(3 dx)  3
9x 2
3
u 2  25
9 x 2  25
2
du

3
ln
3
x

9
x

25

C.
x
u2
(13) Make the trigonometric substitution x  sec   dx  sec  tan  d and
x 2  1  tan  

x2 1
dx 
x
  sec   1 d  tan     C 

(tan  )(sec  tan  d )

sec 
 tan  d 
2
2
x
x  1  arc sec x  C .
x 1
2
2

1
(14) Make the trigonometric substitution x  sin   dx  cos  d and 1  x 2 
cos  

x3
dx 

sin 3 
cos  d  
cos 
1 x2
1
2
sin 3  d   sin 2  cos   cos   C 
3
3
1
2
 x2 1 x2 
1 x2  C .
3
3

x
1

1 x2
(15) Make the trigonometric substitution x  tan   dx  sec 2  d and
13
x 2 1 
sec  

ln

1
x 1
2
dx 

1
(sec 2  d ) 
sec 
sec  d  ln sec   tan   C 
x2 1
x

x2 1  x  C .
1
(16) Make the trigonometric substitution x  tan   dx  sec 2 x dx and 1  x 2 
1
sec 2 
dx

d 
(1  x 2 ) 2
sec 4 
1
1
cos 2 d    sin  cos   C 
2
2
1
1
x  1 

C 
arctan x  
2
2  1  x 2  1  x 2 
1
1 x 
arctan x  
C.
2
2 1 x2 
sec 2  



14
1 x2
x

1